Galois Group of x^4+px+p (14.6.15)

Dummit and Foote Abstract Algebra, section 14.6, exercise 15:

Prove that the polynomial $f(x)=x^4+px+p∈ℚ[x]$ is irreducible for every prime $p$ and for $p \neq 3,5$ has Galois group $S_4$. Prove the Galois group for $p=3$ is dihedral of order 8 and for $p=5$ is cyclic of order $4$.

Proof: We check for rational roots to see if it has a linear factor; $f(-p)=p(p^3-p+1)≠0$, $f(-1)=1$, and $f(1),f(p) > 0$, so we check to see if $f(x)$ splits into two irreducible quadratics. $$f(x)=(x^2+ax+b)(x^2+cx+d)$$ Assume without loss that $d= \pm 1$ and $b= \pm p$. Then assuming $d=1$ and $b=p$ we have by the coefficient of $x$ the relation $a-ap=p$ so $a(1-p)=p$ and testing $a=-1,1,p,-p$ all yield immediate contradictions except where $a=-p$ implying $p=2$, but then $c=2$ and we evidently have $x^4+2x+2≠(x^2-2x+2)(x^2+2x+1)$. The case when $d=-1$ and $b=-p$ is parallel.

Hence $f(x)$ is irreducible over $ℚ$. Now, we note that the resolvent cubic $g(x)$ of $f(x)$ is $x^3-4px+p^2$. To test irreducibility it suffices to check for rational roots. We see $g(p)=p^2(p-3)$ so $p=3$ is a special case here. We see $g(1)=1-4p+p^2≡1~\text{mod p}$ so $g(1)≠0$. We see $g(-1)=-1+4p+p^2≠0$. Finally, $g(-p)=-p^2(p-5)$ so $p=5$ is also a special case. Otherwise, the the resolvent cubic is irreducible. Now, we examine the determinant $D=p^3(256-27p)$. When $p≠2$ we see $p \not \mid 256-27p$ so $D$ has a nonsquare term in its prime factorization. When $p=2$ we have $D=1616=2^4·101$ which is also not a square. Hence for $p≠3,5$ the Galois group is $S_4$.

We further examine $x^4+3x+3$ when $p=3$. As we saw $g(x)=x^3-12x+9=(x-3)(x^2+3x-3)$. Since $3^2+4·3=21$ is not a square we see the quadratic is irreducible. We see $D=4725$ so we examine if $x^4+3x+3$ factors over $ℚ(\sqrt{4725})$. We see the latter is contained in $ℝ$, so it suffices to show $x^4+3x+3$ has no real roots by showing it evaluates positive for all real values. We calculate its derivative $4x^3+3$ is zero when $x=-\sqrt[3]{\dfrac{3}{4}}$, which must be the local minimum as the second derivative $12x^2$ is positive. But assuming $f(-\sqrt[3]{\dfrac{3}{4}})≤0$ implies $\dfrac{3}{4}≥\sqrt[3]{\dfrac{3}{4}}$, a contradiction. Hence $x^4+3x+3$ remains irreducible and the Galois group is dihedral.

We further examine $x^4+5x+5$ when $p=5$. As we saw $g(x)=x^3-20x+25=(x+5)(x^2-5x+5)$. Since $5^2-5·4=5$ is not a square we see the quadratic is irreducible. We see $D=15125=5^3·11^2$ so $ℚ(\sqrt{D})=ℚ(\sqrt{5})$.